Skip to content

Navigation Menu

Sign in
Appearance settings

Search code, repositories, users, issues, pull requests...

Provide feedback

We read every piece of feedback, and take your input very seriously.

Saved searches

Use saved searches to filter your results more quickly
Sign up
Appearance settings

feat: add solutions to lc problem: No.0760 #4806

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Merged
yanglbme merged 1 commit into main from dev
Oct 29, 2025
Merged
Show file tree
Hide file tree
Changes from all commits
Commits
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
77 changes: 63 additions & 14 deletions solution/0700-0799/0760.Find Anagram Mappings/README.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -58,7 +58,11 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一:哈希表

我们用一个哈希表 $\textit{d}$ 来存储数组 $\textit{nums2}$ 中每个元素及其对应的下标。然后我们遍历数组 $\textit{nums1},ドル对于每个元素 $\textit{nums1}[i],ドル我们从哈希表 $\textit{d}$ 中获取其对应的下标并存入结果数组中。

时间复杂度 $O(n),ドル空间复杂度 $O(n),ドル其中 $n$ 是数组的长度。

<!-- tabs:start -->

Expand All @@ -67,29 +71,74 @@ tags:
```python
class Solution:
def anagramMappings(self, nums1: List[int], nums2: List[int]) -> List[int]:
mapper = defaultdict(set)
for i, num in enumerate(nums2):
mapper[num].add(i)
return [mapper[num].pop() for num in nums1]
d = {x: i for i, x in enumerate(nums2)}
return [d[x] for x in nums1]
```

#### Java

```java
class Solution {
public int[] anagramMappings(int[] nums1, int[] nums2) {
Map<Integer, Set<Integer>> map = new HashMap<>();
for (int i = 0; i < nums2.length; ++i) {
map.computeIfAbsent(nums2[i], k -> new HashSet<>()).add(i);
int n = nums1.length;
Map<Integer, Integer> d = new HashMap<>(n);
for (int i = 0; i < n; ++i) {
d.put(nums2[i], i);
}
int[] ans = new int[n];
for (int i = 0; i < n; ++i) {
ans[i] = d.get(nums1[i]);
}
int[] res = new int[nums1.length];
for (int i = 0; i < nums1.length; ++i) {
int idx = map.get(nums1[i]).iterator().next();
res[i] = idx;
map.get(nums1[i]).remove(idx);
return ans;
}
}
```

#### C++

```cpp
class Solution {
public:
vector<int> anagramMappings(vector<int>& nums1, vector<int>& nums2) {
int n = nums1.size();
unordered_map<int, int> d;
for (int i = 0; i < n; ++i) {
d[nums2[i]] = i;
}
return res;
vector<int> ans;
for (int x : nums1) {
ans.push_back(d[x]);
}
return ans;
}
};
```

#### Go

```go
func anagramMappings(nums1 []int, nums2 []int) []int {
d := map[int]int{}
for i, x := range nums2 {
d[x] = i
}
ans := make([]int, len(nums1))
for i, x := range nums1 {
ans[i] = d[x]
}
return ans
}
```

#### TypeScript

```ts
function anagramMappings(nums1: number[], nums2: number[]): number[] {
const d: Map<number, number> = new Map();
for (let i = 0; i < nums2.length; ++i) {
d.set(nums2[i], i);
}
return nums1.map(num => d.get(num)!);
}
```

Expand Down
77 changes: 63 additions & 14 deletions solution/0700-0799/0760.Find Anagram Mappings/README_EN.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -55,7 +55,11 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: Hash Table

We use a hash table $\textit{d}$ to store each element of the array $\textit{nums2}$ and its corresponding index. Then we iterate through the array $\textit{nums1},ドル and for each element $\textit{nums1}[i],ドル we retrieve its corresponding index from the hash table $\textit{d}$ and store it in the result array.

The time complexity is $O(n)$ and the space complexity is $O(n),ドル where $n$ is the length of the array.

<!-- tabs:start -->

Expand All @@ -64,29 +68,74 @@ tags:
```python
class Solution:
def anagramMappings(self, nums1: List[int], nums2: List[int]) -> List[int]:
mapper = defaultdict(set)
for i, num in enumerate(nums2):
mapper[num].add(i)
return [mapper[num].pop() for num in nums1]
d = {x: i for i, x in enumerate(nums2)}
return [d[x] for x in nums1]
```

#### Java

```java
class Solution {
public int[] anagramMappings(int[] nums1, int[] nums2) {
Map<Integer, Set<Integer>> map = new HashMap<>();
for (int i = 0; i < nums2.length; ++i) {
map.computeIfAbsent(nums2[i], k -> new HashSet<>()).add(i);
int n = nums1.length;
Map<Integer, Integer> d = new HashMap<>(n);
for (int i = 0; i < n; ++i) {
d.put(nums2[i], i);
}
int[] ans = new int[n];
for (int i = 0; i < n; ++i) {
ans[i] = d.get(nums1[i]);
}
int[] res = new int[nums1.length];
for (int i = 0; i < nums1.length; ++i) {
int idx = map.get(nums1[i]).iterator().next();
res[i] = idx;
map.get(nums1[i]).remove(idx);
return ans;
}
}
```

#### C++

```cpp
class Solution {
public:
vector<int> anagramMappings(vector<int>& nums1, vector<int>& nums2) {
int n = nums1.size();
unordered_map<int, int> d;
for (int i = 0; i < n; ++i) {
d[nums2[i]] = i;
}
return res;
vector<int> ans;
for (int x : nums1) {
ans.push_back(d[x]);
}
return ans;
}
};
```

#### Go

```go
func anagramMappings(nums1 []int, nums2 []int) []int {
d := map[int]int{}
for i, x := range nums2 {
d[x] = i
}
ans := make([]int, len(nums1))
for i, x := range nums1 {
ans[i] = d[x]
}
return ans
}
```

#### TypeScript

```ts
function anagramMappings(nums1: number[], nums2: number[]): number[] {
const d: Map<number, number> = new Map();
for (let i = 0; i < nums2.length; ++i) {
d.set(nums2[i], i);
}
return nums1.map(num => d.get(num)!);
}
```

Expand Down
15 changes: 15 additions & 0 deletions solution/0700-0799/0760.Find Anagram Mappings/Solution.cpp
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,15 @@
class Solution {
public:
vector<int> anagramMappings(vector<int>& nums1, vector<int>& nums2) {
int n = nums1.size();
unordered_map<int, int> d;
for (int i = 0; i < n; ++i) {
d[nums2[i]] = i;
}
vector<int> ans;
for (int x : nums1) {
ans.push_back(d[x]);
}
return ans;
}
};
11 changes: 11 additions & 0 deletions solution/0700-0799/0760.Find Anagram Mappings/Solution.go
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,11 @@
func anagramMappings(nums1 []int, nums2 []int) []int {
d := map[int]int{}
for i, x := range nums2 {
d[x] = i
}
ans := make([]int, len(nums1))
for i, x := range nums1 {
ans[i] = d[x]
}
return ans
}
19 changes: 9 additions & 10 deletions solution/0700-0799/0760.Find Anagram Mappings/Solution.java
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -1,15 +1,14 @@
class Solution {
public int[] anagramMappings(int[] nums1, int[] nums2) {
Map<Integer, Set<Integer>> map = new HashMap<>();
for (int i = 0; i < nums2.length; ++i) {
map.computeIfAbsent(nums2[i], k -> new HashSet<>()).add(i);
int n = nums1.length;
Map<Integer, Integer> d = new HashMap<>(n);
for (int i = 0; i < n; ++i) {
d.put(nums2[i], i);
}
int[] res = new int[nums1.length];
for (int i = 0; i < nums1.length; ++i) {
int idx = map.get(nums1[i]).iterator().next();
res[i] = idx;
map.get(nums1[i]).remove(idx);
int[] ans = new int[n];
for (int i = 0; i < n; ++i) {
ans[i] = d.get(nums1[i]);
}
return res;
return ans;
}
}
}
6 changes: 2 additions & 4 deletions solution/0700-0799/0760.Find Anagram Mappings/Solution.py
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -1,6 +1,4 @@
class Solution:
def anagramMappings(self, nums1: List[int], nums2: List[int]) -> List[int]:
mapper = defaultdict(set)
for i, num in enumerate(nums2):
mapper[num].add(i)
return [mapper[num].pop() for num in nums1]
d = {x: i for i, x in enumerate(nums2)}
return [d[x] for x in nums1]
7 changes: 7 additions & 0 deletions solution/0700-0799/0760.Find Anagram Mappings/Solution.ts
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,7 @@
function anagramMappings(nums1: number[], nums2: number[]): number[] {
const d: Map<number, number> = new Map();
for (let i = 0; i < nums2.length; ++i) {
d.set(nums2[i], i);
}
return nums1.map(num => d.get(num)!);
}

AltStyle によって変換されたページ (->オリジナル) /